RD Sharma solution class 8 chapter 1 Rational Numbers Exercise 1.8

Exercise 1.8

Page-1.43

Question 1:

Find a rational number between −3 and 1.

Answer 1:

$R\mathrm{ational} \mathrm{number} \mathrm{between} -3 \mathrm{and} 1 = \frac{-3+1}{2}=\frac{-2}{2}=-1$

Question 2:

 Find any five rational numbers less than 2.

Answer 2:

$$\begin{gathered} \mathrm{We} \mathrm{can} \mathrm{write}: \\ 2=\frac{2}{1}=\frac{2\times 5}{1\times 5}=\frac{10}{5} \\ \mathrm{Integers} \mathrm{less} \mathrm{than} 10 \mathrm{are} 0, 1, 2, 3, 4, 5 ... 9. \\ \mathrm{Hence}, \mathrm{five} \mathrm{rational} \mathrm{numbers} \mathrm{less} \mathrm{than} 2 \mathrm{are} \frac{0}{5},\frac{1}{5},\frac{2}{5},\frac{3}{5} \mathrm{and}\frac{4}{5} . \end{gathered}$$

Question 3:

Find two rational numbers between $\frac{-2}{9} \mathrm{and} \frac{5}{9}.$

Answer 3:

$$\begin{gathered} Since both the fractions (\frac{-2}{9} and \frac{5}{9}) have the same denominator, the integers between the numerators(-2 and 5) are -1, 0, 1, 2, 3, 4. \\ Hence, two rational numbers between \frac{-2}{9} and \frac{5}{9} are \frac{0}{9} or 0 and \frac{1}{9}. \end{gathered}$$

Question 4:

Find two rational numbers between $\frac{1}{5} \mathrm{and} \frac{1}{2}.$

Answer 4:

$$\begin{gathered} R\mathrm{ational} \mathrm{number} \mathrm{between} \frac{1}{5} \mathrm{and} \frac{1}{2}=\frac{(\frac{1}{5}+\frac{1}{2})}{2}=\frac{{\displaystyle \frac{2+5}{10}}}{2}=\frac{7}{20} \\ R\mathrm{ational} \mathrm{number} \mathrm{between} \frac{1}{5} \mathrm{and} \frac{7}{20}=\frac{(\frac{1}{5}+\frac{7}{20})}{2}=\frac{{\displaystyle \frac{4+7}{20}}}{2}=\frac{11}{40} \\ \mathrm{Therefore}, \mathrm{two} \mathrm{rational} \mathrm{numbers} \mathrm{between}\frac{1}{5} \mathrm{and} \frac{1}{2} \mathrm{are} \frac{7}{20} \mathrm{and} \frac{11}{40}. \end{gathered}$$

Question 5:

Find ten rational numbers between $\frac{1}{4} \mathrm{and} \frac{1}{2}.$

Answer 5:

$$\begin{gathered} The L.C.M of the denominators (2 and 4) is 4. \\ So, we can write \frac{1}{4} as it is. \\ Also, \frac{1}{2}=\frac{1\times 2}{2\times 2}=\frac{2}{4} \\ As the integers between the numerators 1 and 2 of both the fractions are not sufficient, we will multiply the fractions by 20. \\ \therefore \frac{1}{4}=\frac{1\times 20}{4\times 20}=\frac{20}{80} \\ \frac{2}{4}=\frac{2\times 20}{4\times 20}=\frac{40}{80} \\ Between 20 and 40, there are 19 integers. They are 21, 22, 23, 24, 25, 26, 27....39, 40. \\ Thus, \frac{21}{40},\frac{22}{40},\frac{23}{40},\frac{24}{40},\frac{25}{40},...................\frac{38}{40} and \frac{39}{40} are the fractions. \\ We can take any 10 of these. \end{gathered}$$

Question 6:

Find ten rational numbers between $\frac{-2}{5} \mathrm{and} \frac{1}{2}.$

Answer 6:

$$\begin{gathered} \mathrm{L}.\mathrm{C}.\mathrm{M} \mathrm{of} \mathrm{the} \mathrm{denominator}s (2 \mathrm{and} 5) \mathrm{is} 10. \\ We \mathrm{can} \mathrm{write}: \\ \frac{-2}{5}=\frac{-2\times 2}{5\times 2}=\frac{-4}{10} \\ \mathrm{and} \frac{1}{2}=\frac{1\times 5}{2\times 5}=\frac{5}{10} \\ Since the \mathrm{integers} \mathrm{between} \mathrm{the} \mathrm{numerators} (-4 \mathrm{and} 5 ) \mathrm{of} \mathrm{both} the \mathrm{fractions} are not sufficient, \mathrm{we} will \mathrm{multiply} \mathrm{the} \mathrm{fractions} \mathrm{by} 2. \\ \therefore \frac{-4}{10}=\frac{-4\times 2}{10\times 2}=\frac{-8}{20} \\ \frac{5}{10}=\frac{5\times 2}{10\times 2}=\frac{10}{20} \\ There are 17 integers b\mathrm{etween} -8 \mathrm{and} 10, which are -7,-6,-5,-4...................8, 9. \\ These can be written as: \\ \frac{-7}{20},\frac{-6}{20},\frac{-5}{20},\frac{-4}{20},\frac{-3}{20},...................\frac{8}{20} \mathrm{and} \frac{9}{20} \\ \mathrm{We} \mathrm{can} \mathrm{take} \mathrm{any} 10 \mathrm{of} \mathrm{these}. \end{gathered}$$

Question 7:

Find ten rational numbers between $\frac{3}{5} \mathrm{and} \frac{3}{4}.$

Answer 7:

$$\begin{gathered} The L.C.M of the denominators 5 and 4 of both the fractions is 20. \\ We can write: \\ \frac{3}{5}=\frac{3\times 4}{5\times 4}=\frac{12}{20} \\ \frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20} \\ Since the integers between the numerators 12 and 15 are not sufficient, we will multiply both the fractions by 5. \\ \frac{12}{20}=\frac{12\times 5}{20\times 5}=\frac{60}{100} \\ \frac{15}{20}=\frac{15\times 5}{20\times 5}=\frac{75}{100} \\ There are 14 integers between 60 and 75. They are 61, 62, 63.......73 and 74. \\ Therefore, \frac{60}{100},\frac{61}{100},\frac{62}{100}..........\frac{73}{100} and\frac{74}{100} are the 14 fractions. \\ We can take any 10 of these. \end{gathered}$$